Series Convergence: Does \sum_{k=1}^{\infty }{a_k}^{5/4} Necessarily Converge?

[Screwdriver]

Homework Statement

Say that

\sum_{k=1}^{\infty }a_k

converges and has positive terms. Does the following necessarily converge?

\sum_{k=1}^{\infty }{a_k}^{5/4}

Homework Equations

If it necessarily converges, a proof is required, if not, a counter-example is required.

The Attempt at a Solution

I suspect that it diverges, so I tried an arbitrary geometric series, but it didn't work:

\sum_{k=1}^{\infty }({\frac{1}{k^n}})^{5/4} = \sum_{k=1}^{\infty }({\frac{1}{k^{5n/4}}})

But n must be greater than 1 since a_k must converge, and anything greater than 1 times 5/4 is still greater than one, so nothing's been proven.

Alternatively, I could suspect that it will necessarily converge, and use the comparison test:

{a_k}^{5/4}{\leq }^?a_k
{a_k}^{1/4}{\leq }^?1
{a_k}{\leq }^?1

But that's not going to happen in general.

 

[l'Hôpital]

Think about it what means for

<br /> \sum_{k=1}^{\infty }a_k<br />

To converge. What does that imply about a_k? And with that knowledge, what does that entail about a_k ^(c) for c > 1 with relation to a_k?

 

[Screwdriver]

If

\sum_{k=1}^{\infty }a_k

converges, then the terms of the series a_k are approaching zero since

\lim_{x \to 0 }x^c = 0

for c&gt;1. Therefore, the terms of {a_k}^{5/4} are also approaching zero, and they must converge.

Right?

 

[l'Hôpital]

You have the right idea, except you got to use a slightly different argument. After all, if a_k = 1/k, the terms still go to zero, and it all fails.

Try comparison.

 

[Screwdriver]

But a_k \neq \frac{1}{k} because it was given that the series converges. Should I do something like

{a_k}^{c}\leq^? a_k

{a_k}^{c-1}\leq^? 1

And then argue that this will be true for c > 1 (which is required) because a_k converges, so the terms must be approaching zero so they will be less than 1?